| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Differentiation of trigonometric composites |
| Difficulty | Moderate -0.3 This is a straightforward differentiation question testing standard techniques: chain rule for (a), product rule for (b), and quotient rule for (c). While it requires knowledge of multiple differentiation rules and trigonometric derivatives, these are core C3 skills with no problem-solving or novel insight required. The 9 total marks reflect routine application rather than difficulty. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(= -\cos e^2 \cdot x \times 2x = -2x\cos e^2 x\) | M1 A1 | |
| (b) \(= 2x \times e^x + x^2 \times (-e^{-x}) = xe^{-i}(2-x)\) | M1 A2 | |
| (c) \(= \frac{\cos x(3+2\cos x) - \sin x(-2\sin x)}{(3+2\cos x)^2}\) | M1 A1 | |
| \(= \frac{3\cos x + 2\cos^2 x + 2\sin^2 x}{(3+2\cos x)^2} = \frac{3\cos x + 2}{(3+2\cos x)^2}\) | M1 A1 | (9) |
(a) $= -\cos e^2 \cdot x \times 2x = -2x\cos e^2 x$ | M1 A1 |
(b) $= 2x \times e^x + x^2 \times (-e^{-x}) = xe^{-i}(2-x)$ | M1 A2 |
(c) $= \frac{\cos x(3+2\cos x) - \sin x(-2\sin x)}{(3+2\cos x)^2}$ | M1 A1 |
$= \frac{3\cos x + 2\cos^2 x + 2\sin^2 x}{(3+2\cos x)^2} = \frac{3\cos x + 2}{(3+2\cos x)^2}$ | M1 A1 | **(9)**Differentiate each of the following with respect to $x$ and simplify your answers.
\begin{enumerate}[label=(\alph*)]
\item $\cot x^2$ [2]
\item $x^2 e^{-x}$ [3]
\item $\frac{\sin x}{3 + 2\cos x}$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q3 [9]}}